In: Math
For the following matrices, first find a basis for the column space of the matrix. Then use the Gram-Schmidt process to find an orthogonal basis for the column space. Finally, scale the vectors of the orthogonal basis to find an orthonormal basis for the column space.
(a) [1 1 1, 1 0 2, 3 1 0, 0 0 4 ] b) [?1 6 6, 3 ?8 3, 1 ?2 6, 1 ?4 ?3 ]
We presume that the triplets mentioned are the columns of the matrix.
(a). Let A =
1 |
1 |
3 |
0 |
1 |
0 |
1 |
0 |
1 |
2 |
0 |
4 |
The RREF of A is
1 |
0 |
0 |
4/5 |
0 |
1 |
0 |
8/5 |
0 |
0 |
1 |
-4/5 |
Thus, a basis for col(A) is { v1,v2,v3} = { (1,1,1)T,(1,0,2)T,(3,1,0)T} .
Let u1=v1=(1,1,1)T,u2=v2–proju1(v2)=v2–[(v2.u1)/(u1.u1)]u1=v2–[(1+0+2)/(1+1+1)]u1=(1,0,2)T-(1,1,1)T= (0,-1, 1)Tand u3=v3–proju1(v3)=v3–[(v3.u1)/(u1.u1)]u1–[(v3.u2)/(u2.u2)]u2=v3–[(3+1+0)/(1+1+1)]u1-[(0-1+0)/ (0+1+1)]u2 = (3,1,0)T –(4/3)( 1,1,1)T -(1/2)( 0,-1, 1)T= (5/3,-5/6,-5/6)T. Then { u1,u2,u3}= {(1,1,1)T, (0,-1, 1)T, (5/3,-5/6,-5/6)T } is an orthogonal basis for col(A).
Further, let e1 = u1/||u1|| =(1/?3,1?3,1/?3)T e2 = u2/||u2|| =(0,-1?2, 1/?2)Tand e3 = u3/||u3|| = (5/3,-5/6,-5/6)T= (?2/?3, 1/?6,-1/?6)T. Then {e1,e2,e3} = {(1/?3,1?3,1/?3)T, (0,-1?2, 1/?2)T, (?2/?3, 1/?6,-1/?6)T} is an orthonormal basis for col(A).
Please post the other question again.